\(\left\{{}\begin{matrix}x^2=y+6\\y^2=x+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y=6\\y^2-x=6\end{matrix}\right.\)
\(\Rightarrow x^2-y=y^2-x\)
\(\Leftrightarrow x^2-y-y^2+x=0\)
\(\Leftrightarrow\left(x^2-y^2\right)+\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=-1-x\end{matrix}\right.\)
+ Với x = y, ta có:
\(x^2-x=6\) \(\Leftrightarrow\left[{}\begin{matrix}x=y=3\\x=y=-2\end{matrix}\right.\)
+ Với y = -1 - x, ta có: x2 + x - 5 = 0
Còn lại bạn làm nốt nhé!
\(\left\{{}\begin{matrix}x^2=y+6\\y^2=x+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=y-x\\x^2=y+6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y\right)+x-y=0\\x^2-y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+1\right)=0\\x^2-y-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=0\\x+y+1=0\end{matrix}\right.\\x^2-y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-x-6=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1-x\\x^2+x-5=0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1-x\\\left(x+\frac{1}{2}\right)^2=\frac{19}{4}\end{matrix}\right.\end{matrix}\right.\) ... tự giải nốt
